How to find the following condition on the domain of conjugate function of negative geometric mean?

Question

The conjugate function of the negative geometric mean ($f(x)=-(\prod_{i}^{n} x_i)^{1/n}$ over $R^n_{++}$) is given as follows $$f^*(y)=\sup_{x}\left(x^Ty+(\prod_{i}^{n} x_i)^{1/n}\right).$$ I know that the domain of the function must satisfy $y\preceq 0$. But, for this function how to get the following condition on the domain of the conjugate function $$\left(\prod_{i=1}^n |y_i|\right)^{1/n}> \frac{1}{n}. $$ Thanks in advance.

Answer 1

This is not an answer to the original question but rather an answer to the original poster's comment directed at me. (Placed as answer for space reasons.)

Suppose that $n=1$.
Then $$f^*(y)=\sup_{x>0}\big(y\cdot x + x\big) = \sup_{x>0}(y+1)x.$$ Now let us evaluate this supremum.
Case 1: If $y>-1$, i.e., $y+1>0$, then $f^*(y)\geq (y+1)x\to+\infty$ as $x\to+\infty$.
Case 2: If $y=-1$, then $y+1=0$ and clearly $f^*(y)=0$.
Case 3: If $y<-1$, then $y+1<0$. Hence $(y+1)x<0$ when $x>0$, and this implies $f^*(y)\leq 0$. However, if $x\to 0^+$, then $(y+1)x\to 0^-$ and so $f^*(y)=0$.

Altogether, $$f^*(y) = \begin{cases} 0, &\text{if $y\leq -1$;}\\ +\infty, &\text{if $y>-1$} \end{cases}.$$

So, for $n=1$, it is not true that $|y|>1$ when $y$ is in the domain of $f^*$.